--- title: "Dielectric function" firstLetter: "D" publishDate: 2022-01-24 categories: - Physics - Electromagnetism - Quantum mechanics date: 2022-01-20T22:04:13+01:00 draft: false markup: pandoc --- # Dielectric function The **dielectric function** or **relative permittivity** $\varepsilon_r$ is a measure of how strongly a given medium counteracts [electric fields](/know/concept/electric-field/) compared to a vacuum. Let $\vb{D}$ be the applied external field, and $\vb{E}$ the effective field inside the material: $$\begin{aligned} \boxed{ \vb{D} = \varepsilon_0 \varepsilon_r \vb{E} } \end{aligned}$$ If $\varepsilon_r$ is large, then $\vb{D}$ is strongly suppressed, because the material's electrons and nuclei move to create an opposing field. In order for $\varepsilon_r$ to be well defined, we only consider linear media, where the induced polarization $\vb{P}$ is proportional to $\vb{E}$. We would like to find an alternative definition of $\varepsilon_r$. Consider that the usual electric fields $\vb{E}$, $\vb{D}$, and $\vb{P}$ can each be written as the gradient of an electrostatic potential like so, where $\Phi_\mathrm{tot}$, $\Phi_\mathrm{ext}$ and $\Phi_\mathrm{ind}$ are the total, external and induced potentials, respectively: $$\begin{aligned} \vb{E} = -\nabla \Phi_\mathrm{tot} \qquad \qquad \vb{D} = - \varepsilon_0 \nabla \Phi_\mathrm{ext} \qquad \qquad \vb{P} = \varepsilon_0 \nabla \Phi_\mathrm{ind} \end{aligned}$$ Such that $\Phi_\mathrm{tot} = \Phi_\mathrm{ext} + \Phi_\mathrm{ind}$. Inserting this into $\vb{D} = \varepsilon_0 \varepsilon_r \vb{E}$ then suggests defining: $$\begin{aligned} \boxed{ \varepsilon_r \equiv \frac{\Phi_\mathrm{ext}}{\Phi_\mathrm{tot}} } \end{aligned}$$ ## From induced charge density A common way to calculate $\varepsilon_r$ is from the induced charge density $\rho_\mathrm{ind}$, i.e. the offset caused by the material's particles responding to the field. We start from [Gauss' law](/know/concept/maxwells-equations/) for $\vb{P}$: $$\begin{aligned} \nabla \cdot \vb{P} = \varepsilon_0 \nabla^2 \Phi_\mathrm{ind}(\vb{r}) = - \rho_\mathrm{ind}(\vb{r}) \end{aligned}$$ This is Poisson's equation, which has the following well-known [Fourier transform](/know/concept/fourier-transform/): $$\begin{aligned} \Phi_\mathrm{ind}(\vb{q}) = \frac{\rho_\mathrm{ind}(\vb{q})}{\varepsilon_0 |\vb{q}|^2} = V(\vb{q}) \: \rho_\mathrm{ind}(\vb{q}) \end{aligned}$$ Where $V(\vb{q})$ represents Coulomb interactions, and $V(0) = 0$ to ensure overall neutrality: $$\begin{aligned} V(\vb{q}) = \frac{1}{\varepsilon_0 |\vb{q}|^2} \qquad \implies \qquad V(\vb{r} - \vb{r}') = \frac{1}{4 \pi \varepsilon_0 |\vb{r} - \vb{r}'|} \end{aligned}$$ The [convolution theorem](/know/concept/convolution-theorem/) then gives us the solution $\Phi_\mathrm{ind}$ in the $\vb{r}$-domain: $$\begin{aligned} \Phi_\mathrm{ind}(\vb{r}) = (V * \rho_\mathrm{ind})(\vb{r}) = \int_{-\infty}^\infty V(\vb{r} - \vb{r}') \: \rho_\mathrm{ind}(\vb{r}') \dd{\vb{r}'} \end{aligned}$$ To proceed, we need to find an expression for $\rho_\mathrm{ind}$ that is proportional to $\Phi_\mathrm{tot}$ or $\Phi_\mathrm{ext}$, or some linear combination thereof. Such an expression must exist for a linear material. Suppose we can show that $\rho_\mathrm{ind} = C_\mathrm{ext} \Phi_\mathrm{ext}$, for some $C_\mathrm{ext}$, which may depend on $\vb{q}$. Then: $$\begin{aligned} \Phi_\mathrm{tot} = (1 + C_\mathrm{ext} V) \Phi_\mathrm{ext} \quad \implies \quad \boxed{ \varepsilon_r(\vb{q}) = \frac{1}{1 + C_\mathrm{ext}(\vb{q}) V(\vb{q})} } \end{aligned}$$ Similarly, suppose we can show that $\rho_\mathrm{ind} = C_\mathrm{tot} \Phi_\mathrm{tot}$, for some quantity $C_\mathrm{tot}$, then: $$\begin{aligned} \Phi_\mathrm{ext} = (1 - C_\mathrm{tot} V) \Phi_\mathrm{tot} \quad \implies \quad \boxed{ \varepsilon_r(\vb{q}) = 1 - C_\mathrm{tot}(\vb{q}) V(\vb{q}) } \end{aligned}$$ ## References 1. H. Bruus, K. Flensberg, *Many-body quantum theory in condensed matter physics*, 2016, Oxford. 2. M. Fox, *Optical properties of solids*, 2nd edition, Oxford.